1 / 15

Lecture 15

ENGR-1100 Introduction to Engineering Analysis. Lecture 15. Today Lecture Outline. Equilibrium of a rigid body in 3D. F ree B ody D iagrams in 3D Equilibrium equations (6 independent equations in 3D). Resultant of a general 3D system of non-parallel non-coplanar forces. z. z. y. y.

carolyn-foster
Download Presentation

Lecture 15

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ENGR-1100 Introduction to Engineering Analysis Lecture 15

  2. Today Lecture Outline • Equilibrium of a rigid body in 3D. • Free Body Diagrams in 3D • Equilibrium equations (6 independent equations in 3D)

  3. Resultant of a general 3D system of non-parallel non-coplanar forces z z y y x1 xR x y1 x yR The procedure 1) Choose a point O (arbitrary). 2) Replace each force by a force at O and a couple perpendicular to it. 3) Combine the forces at O (vectorially) to obtain the resultant force. 4) Combine the couples at O (vectorially) to obtain the resultant couple

  4. Particle equilibrium Rigid body equilibrium Necessary and sufficient conditions of equilibrium of a rigid body The necessary AND sufficient condition for a rigid body to be in equilibrium is that theresultant forceandcoupleacting on the rigid body must be zero. Generally, a system of forces in 3-D can be reduced to aforce-couple system.

  5. 6 Independent Equations in 3-D

  6. 1. Ball and socket 2. Hinge Three-dimensional reaction at supports and connections

  7. 3. Ball bearing 4. Journal bearing

  8. 6. Smooth pin and bracket 5. Thrust bearing 7. Fixed support

  9. Example P6-80 Bar AC of Fig. P6-80 rests against a smooth surface at end C and is supported at end A with a ball-and-socket joint. The cable at B is attached midway between the ends of the bar. Determine the reactions at supports A and C and the tension in thecable at B.

  10. Z Az Ay Ax TBD FBE x y C =388.1 i - 388.1 j - 582.1 k N =TBD(-0.3714 i +0.7428 j - 0.5571 k N

  11. Z Az Ay Ax TBD FBE x y C From a free-body diagram on the bar: The moment equilibrium: = (0.8 i +0.4 j-1.2k)x(388.1i -388.1 j -582.1 k ) +(1.6 j - 1.2 k ) x (-0.3714 TBDi + 0.7428 TBDj -0.5571 TBDk ) +(0.8 i + 1.6 j - 1.2 k ) x (Czk) = ( 1.6 Cz - 698.56) i + (0.4457TBD - 0.8 Cz) j + (0.5942TBD - 465.72) k = 0

  12. Z Az Ay Ax TBD FBE x y C For which: Cz = 436.6 N C = 437 k N TBD= 783.7 N TBD= -291.1 i + 582.2 j -436.7 k N The force equilibrium: = (Axi + Ayj + Azk )+ (388.1i -388.1 j -582.1 k ) - (291.1 i +582.2 j –436.7k ) +436.6 k =0 (Ax +97)i + (Ay +194.1)j + (Az-582.2)k =0

  13. (Ax +97)i + (Ay +194.1)j + (Az-582.2)k =0 Ax =-97 N Ay =-194.1 N Az=582.2 Or: Question: What does it mean to have 5 unknown forces vs, 6 equations?

  14. Class Assignment: Exercise set 6-79 please submit to TA at the end of the lecture Bar AB is used to support an 850-lb load as shown in Fig. P6-79. End A of the bar is supported with a ball-and-socket joint. End B of the bar is supported with two cables. Determine the components of the reaction at support A and the tensions in the two cables. Answer: TBC= -1046 j lb; TBD= -523 i lb; A= 523 i + 1046 j + 850 k lb

  15. Class Assignment: Exercise set 6-87 please submit to TA at the end of the lecture The plate shown in Fig. P6-87 weighs 150 lb and is supported in a horizontal position by two hinges and a cable, The hinges have been properly aligned; therefore, they exert only force reactions on the plate. Assume that the hinge at B resists any force along the axis of the hinge pins. Determine the reactions at supports A and B and the tension in the cable. Answer: TC= 83i –91.67 j + 75 k lb; A= 107j +50 k lb; B=- 83 i –15.31j + 25 k lb

More Related